# Impulse Response of Causal Systems

1. Jan 26, 2013

### AcidRainLiTE

I am reading "Linear System Theory and Design" by Chen and he says (in what follows g(t,tau) is the impulse response function):

If a system is causal, the output will not appear before an input is applied. Thus we have Causal <==> g(t,tau) = 0 for t < tau.​

However, this seems incorrect to me. For a given tau, g(t,tau) represents the output of the system in response to a delta spike centered at t= tau. So, for t < tau, the input stimulating g(t,tau) is 0. The author then concludes from this that the output for t < tau must also be zero. But this is false. Consider for instance a system consisting of a logical NOT gate where your the output is the logical NOT of your input. g(t,tau) would not be zero for t<tau for this system. What am I missing?

2. Jan 26, 2013

### AlephZero

Usually, a NOT gate would mean you were talikng about digital signals, not anlog (continuously variable) ones.

Even if you extend your idea to an "analog NOT gate" where the "output = 1 - input" or sometning similar, that is still not a linear system.

For a linear system, if the input is always zero, the output must also be always zero.

3. Jan 26, 2013

### AcidRainLiTE

Do you mean that, for a linear system, if the input and state variables are zero then the output must also be zero?

My textbook defines a linear system as a system which satisfies the following (x denotes state variable, u input, y output):

Given

$$\begin{array}{l l} x_1(t_0) \\ u_1(t), \quad t \geq t_0 \\ \end{array} \} \rightarrow y_1(t), \quad t \geq t_0$$

and

$$\begin{array}{l l} x_2(t_0) \\ u_2(t), \quad t \geq t_0 \\ \end{array} \} \rightarrow y_2(t), \quad t \geq t_0$$

we have that for any $\alpha_1, \alpha_2$

$$\begin{array}{l l} \alpha_1 x_1(t_0) + \alpha_2 x_2(t_0) \\ \alpha_1 u_1(t) + \alpha_2 u_2(t), \quad t \geq t_0 \\ \end{array} \} \rightarrow \alpha_1 y_1(t) + \alpha_2 y_2(t), \quad t \geq t_0$$

I can see how you can deduce from this condition that, if both $x(t_0) = 0$ and $u(t_0) = 0$ then $y(t_0) = 0$. But I cannot see how to do it without the condition on the state variables.

4. Jan 27, 2013

### AlephZero

Yes - sorry if the lack of precision confused you.